Instability Mechanism of Pillar Burst in Asymmetric Mining Based on Cusp Catastrophe Model

Abstract

Pillars are often reserved asymmetrically in the mining process. The roof deflection curve under non-equal span conditions of adjacent stopes is derived by considering the roof-pillar system as a rock beam-pillar model. The pillar instability condition under asymmetric mining is determined based on instability theory and cusp catastrophe theory. Pillar burst represents the equilibrium stability of the roof-pillar system. The pillar failure may be in a violent manner or a gentle manner, depending on the post-peak stiffness ratio of the roof-pillar system. By calculating the factor of safety (FOS) and roof-pillar stiffness ratio K, the pillar stability with different stope spans can be evaluated. The theoretical results are validated by comparison with a case study and numerical simulation. When the stope spans are not equal, the pillar is affected by small-eccentric compression. Four pillar failure patterns under eccentric compression are proposed and explained. The main factors affecting pillar burst appear to include the geometric parameters and mechanical properties of the roof-pillar system. It is difficult to change the mechanical properties, but the stiffness ratio K can be increased by improving the geometric parameters, so as to minimize the burst tendency. Once K < 1 and the critical compression failure load is reached, the pillar on the larger stope span side fails first, and then, the whole pillar loses its stability. Considering the external work during the pillar unstable failure, the rockburst energy index is optimized.

Appendix

The singular function method (Bedford and Liechti 2020) is introduced here as it was used to calculate the deflection equation of the rock beam shown in Fig. 1. There is a function family f_n (x):

$$f_{n} \left( x \right) = \left\langle {x - s} \right\rangle^{n} = \left\{ {\begin{array}{*{20}l} {\left( {x - s} \right)^{n} } \hfill & {x \ge s} \hfill \\ 0 \hfill & {x < s} \hfill \\ \end{array} } \right.,$$

(32)

which is a singular function. Where s is an arbitrary constant. When n = 0, 1, 2, they are constant, primary, and secondary functions, respectively. The integral and derivative of the singular function are:

$$\int {\left\langle {x - s} \right\rangle^{n} {\text{d}}x} = \frac{1}{n + 1}\left\langle {x - s} \right\rangle^{n + 1} + C\quad n \ge 0$$

(33)

$$\frac{d}{{{\text{d}}x}}\left\langle {x - s} \right\rangle^{n} = \left\{ {\begin{array}{*{20}l} 0 \hfill & {n = 0} \hfill \\ {n\left\langle {x - s} \right\rangle^{n - 1} } \hfill & {n \ge 1} \hfill \\ \end{array} } \right..$$

(34)

The mechanical model shown in Fig. 1c is fixed at both ends of the rock beam. If the fixed constraint is removed, the supporting forces R_A, R_B and bending moments M_A, M_B at points A and B are needed to maintain the equilibrium. The horizontal force on the rock beam is ignored, as it has little effect on the rock beam. The generalized forces on the rock beam include uniform load q; pillar support reaction R; bending moment M_C; supporting forces R_A, R_B; and bending moments M_A, M_B. Taking point A as the coordinate origin, the shear force and bending moment equations of the rock beam are as follows:

$$F_{{\text{S}}} = R_{A} + R\left\langle {x - a} \right\rangle^{0} - qx$$

(35)

$$M = R_{A} x - M_{A} + R\left\langle {x - a} \right\rangle^{1} - M_{C} \left\langle {x - a} \right\rangle^{0} - \frac{{qx^{2} }}{2},$$

(36)

where F_S and M are shear force and bending moment, respectively. a is the distance from the pillar center to the left boundary of the UMO. The approximate differential equation of rock beam deflection curve is:

$$E_{{\text{b}}} I_{{\text{b}}} y^{\prime\prime} = - M\left( x \right),$$

(37)

where E_bI_b is the bending stiffness of the beam. y is the rock beam deflection. By substituting Eqs. (36) into (37) and integrating twice, the rotation angle and deflection equation of rock beam are expressed as follows:

$$ \begin{aligned}E_{{\text{b}}} I_{{\text{b}}} \theta &= E_{{\text{b}}} I_{{\text{b}}} y^{\prime} = - \frac{{R_{A} }}{2}x^{2} + M_{A} x - \frac{R}{2}\left\langle {x - a} \right\rangle ^{2} \\ &\quad+ M_{C} \left\langle {x - a} \right\rangle ^{1} + \frac{q}{6}x^{3} + C_{1}\end{aligned} $$

(38)

$$ \begin{aligned}E_{{\text{b}}} I_{{\text{b}}} y &= - \frac{{R_{A} }}{6}x^{3} + \frac{1}{2}M_{A} x^{2} - \frac{R}{6}\left\langle {x - a} \right\rangle ^{3} \\ &\quad+ \frac{1}{2}M_{C} \left\langle {x - a} \right\rangle ^{2} + \frac{q}{{24}}x^{4} + C_{1} x + C_{2} ,\end{aligned} $$

(39)

where θ is the rotation angle of rock beam; C₁ and C₂ are undetermined coefficients. When x = 0, the rotation angle and deflection of rock beam at point A are zero (θ₀ = 0, y₀ = 0). Substituting them into Eqs. (38) and (39), then C₁ = C₂ = 0. Therefore, Eqs. (38) and (39) can be rewritten as follows:

$$E_{{\text{b}}} I_{{\text{b}}} y^{\prime} = E_{{\text{b}}} I_{{\text{b}}} \theta = M_{A} x - \frac{{R_{A} }}{2}x^{2} + M_{C} \left\langle {x - a} \right\rangle - \frac{R}{2}\left\langle {x - a} \right\rangle^{2} { + }\frac{q}{6}x^{3}$$

(40)

$$E_{{\text{b}}} I_{{\text{b}}} y = \frac{{M_{A} }}{2}x^{2} - \frac{{R_{A} }}{6}x^{3} + \frac{{M_{C} }}{2}\left\langle {x - a} \right\rangle^{2} - \frac{R}{6}\left\langle {x - a} \right\rangle^{3} + \frac{q}{24}x^{4} .$$

(41)

The following equation is obtained for the rock beam according to the static equilibrium in y direction:

$$R_{A} + R_{B} = ql - R,$$

(42)

where l is the total span of the rock beam, l = a + b = w_o1 + w_p + w_o2. b is the distance from the pillar center to the right boundary of the UMO. w_o1 and w_o2 are the spans of stope 1 and stope 2, respectively. When x = l, the deflection and rotation angle of rock beam at point B are also zero, i.e., y (x = l) = 0, $y^{\prime}_{{}} (x = l) = 0$. Substituting them into Eqs. (40) and (41), yields the following equations:

$$0 = M_{A} l - \frac{{R_{A} }}{2}l^{2} + M_{C} b - \frac{R}{2}b^{2} + \frac{q}{6}l^{3}$$

(43)

$$0 = \frac{{M_{A} }}{2}l^{2} + \frac{{M_{C} }}{2}b^{2} - \frac{{R_{A} }}{6}l^{3} - \frac{R}{6}b^{3} + \frac{q}{24}l^{4} .$$

(44)

R_A and M_A can be obtained by combining Eqs. (43) and (44) as:

$$R_{A} = \frac{ql}{2} - \left( {1 + \frac{2a}{l}} \right)\frac{{Rb^{2} }}{{l^{2} }} + \frac{{6M_{C} ab}}{{l^{3} }}$$

(45)

$$M_{A} = \frac{{ql^{2} }}{12} - \frac{{Rab^{2} }}{{l^{2} }} + \frac{{M_{C} b\left( {3a - l} \right)}}{{l^{2} }}.$$

(46)

Similarly, we can get R_B and M_B:

$$R_{B} = \frac{ql}{2} - \left( {1 + \frac{2b}{l}} \right)\frac{{Ra^{2} }}{{l^{2} }} - \frac{{6M_{C} ab}}{{l^{3} }}$$

(47)

$$M_{B} = \frac{{ql^{2} }}{12} - \frac{{Ra^{2} b}}{{l^{2} }} - \frac{{M_{C} a\left( {3b - l} \right)}}{{l^{2} }}.$$

(48)

By substituting M_A and R_A into Eq. (41), the deflection equation of rock beam is written as:

$$\begin{aligned} E_{{\text{b}}} I_{{\text{b}}} y & = \left[ {\frac{{ql^{2} }}{12} - \frac{{Rab^{2} }}{{l^{2} }} + \frac{{M_{C} b\left( {3a - l} \right)}}{{l^{2} }}} \right]\frac{{x^{2} }}{2} \\ & \quad - \left[ {\frac{ql}{2} - \left( {1 + \frac{2a}{l}} \right)\frac{{Rb^{2} }}{{l^{2} }} + \frac{{6M_{C} ab}}{{l^{3} }}} \right]\frac{{x^{3} }}{6} \\& \quad + \frac{{M_{C} }}{2}\left\langle {x - a} \right\rangle^{2} - \frac{R}{6}\left\langle {x - a} \right\rangle^{3} + \frac{q}{24}x^{4} . \\ \end{aligned}$$

(49)

Therefore, the equation of rotation angle of rock beam is:

$$\begin{aligned} E_{{\text{b}}} I_{{\text{b}}} y^{\prime} & = E_{{\text{b}}} I_{{\text{b}}} \theta = \left[ {\frac{{ql^{2} }}{12} - \frac{{Rab^{2} }}{{l^{2} }} + \frac{{M_{C} b\left( {3a - l} \right)}}{{l^{2} }}} \right]x \\ & \quad - \left[ {\frac{ql}{2} - \frac{{Rb^{2} \left( {l + 2a} \right)}}{{l^{3} }} + \frac{{6M_{C} ab}}{{l^{3} }}} \right]\frac{{x^{2} }}{2} \\ \quad + M_{C} \left\langle {x - a} \right\rangle - \frac{R}{2}\left\langle {x - a} \right\rangle^{2} + \frac{q}{6}x^{3} . \\ \end{aligned}$$

(50)

R and M_C are still unknown in Eqs. (49) and (50), but they can be determined according to the pillar constitutive equation and the boundary conditions at the bottom of the pillar.